30 
and 
zt C==k tang! 
k Ze ë A Ae 
5 §, = 
Leren 
Wma kh? +k si! —.. 
mV Ea? 
So a solution of the given differential equation containing three 
constants of integration C, m, k has been obtained; we can now 
still investigate in what manner this solution can be transformed 
by means of a suitable choice of the functions §,, §,,§, into the 
most simple form. 
The only condition §,,&,,§, have to satisfy is 
2 A GG, 
So we may put 
5: = A,(a,k°— 1) ’ =, = A,(a,k’* —1) ’ S; = A,(a,k’—1), 
or | 
ib ; Pets ’ §, = 4,6, 
The equations 
at =k ae SA ed 
can be replaced by 
EA =kt , Tb,A27=0, 
and these two are satisfied by putting 
A, Ee A, as 
V(b,-6,)(t-+-6,,) V(b,-b,)(t+6,0,) 
k 
3 
V(b,-b,)(t-+0,b,) V-(b,-b,)(b,-0,)(6,-8,) 
where ¢ is supposed to be determined by the equation 
Ee Kb). 
By eliminating A,, A,, A,, &,,§,,&, out of the solution found above 
we get finally | 
dt 
2-+C—=-hkb,b,b, = 
. | Ta zi b,b,)(¢+-8,5,) 
‚Ze bt) iden) dr + V m? Zath + k sin”! 
Eb, V(b) (4,6) mV Za’ 
By this the differential equation is solved and in this solution the 
symmetry with respect to the independent variables is preserved. 
—ktang— 
